Chapter 4 Trigonometric Functions

Trigonometric Functions
c. θ = 180°= π radians
The terminal side of the angle is on the negative
x-axis. Select the point
P = (–1,0): x =− 1, y = 0, r = 1
Apply the definitions of the cosine and cosecant
functions.
x −1
cos180°= cosπ
= = =−1
r 1
r 1
csc180°= csc π = = , undefined
y 0
d.
3π
θ = 270°= radians
2
The terminal side of the angle is on the negative
y-axis. Select the point
P = (0,–1): x = 0, y = − 1, r = 1
Apply the definitions of the cosine and cosecant
functions.
3π
x 0
cos 270°= cos = = = 0
2 r 1
3π
r 1
csc 270°= csc = = =−1
2 y −1
3. Because sinθ
< 0, θ cannot lie in quadrant I; all the
functions are positive in quadrant I. Furthermore, θ
cannot lie in quadrant II; sinθ is positive in quadrant
II. Thus, with sinθ
< 0, θ lies in quadrant III or
quadrant IV. We are also given that cosθ < 0 .
Because quadrant III is the only quadrant in which
cosine is negative and the sine is negative, we
conclude that θ lies in quadrant III.
4. Because the tangent is negative and the cosine is
negative, θ lies in quadrant II. In quadrant II, x is
negative and y is positive. Thus,
1 y 1
tanθ =− = =
3 x − 3
x =− 3, y = 1
Furthermore,
2 2 2 2
r = x + y = ( − 3) + 1 = 9+ 1=
10
Now that we know x, y, and r, we can find
sinθ and secθ .
y 1 1 10 10
sinθ
= = = ⋅ =
r 10 10 10 10
r 10 10
secθ
= = = −
x −3 3
5. a. Because 210° lies between 180° and 270°, it is
in quadrant III. The reference angle is
θ ′ = 210°− 180°= 30°.
b. Because 7 π 3π
6π
lies between = and
4
2 4
8π
2π = , it is in quadrant IV. The reference
4
7π 8π 7π π
angle is θ′ = 2π
− = − = .
4 4 4 4
c. Because –240° lies between –180° and –270°, it
is in quadrant II. The reference angle is
θ = 240 − 180 = 60°.
d. Because 3.6 lies between π ≈ 3.14 and
3π ≈ 4.71 , it is in quadrant III. The reference
2
angle is θ′ = 3.6 −π
≈ 0.46 .
6. a. 665°− 360°= 305°
This angle is in quadrant IV, thus the reference
angle is θ′ = 360°− 305°= 55°.
b.
c.
15 π 15 8 7
− 2π
= π − π =
π
4 4 4 4
This angle is in quadrant IV, thus the reference
7π 8π 7π π
angle is θ′ = 2π
− = − = .
4 4 4 4
11 11 12
− π + 2⋅ 2π
= − π + π =
π
3 3 3 3
This angle is in quadrant I, thus the reference
π
angle is θ′ = .
3
7. a. 300° lies in quadrant IV. The reference angle is
θ ′ = 360°− 300°= 60°.
b.
3
sin 60°=
2
Because the sine is negative in quadrant IV,
3
sin 300°=− sin 60°=− .
2
5π lies in quadrant III. The reference angle is
4
5 5 4
θ′ = π − π = π − π = π .
4 4 4 4
π
tan = 1
4
Because the tangent is positive in quadrant III,
5π
π
tan =+ tan = 1 .
4 4
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